Question

Prove that for every positive integer$n$, there are$n$ consecutive composite integers. [ Hint: Consider the$n$ consecutive integers starting with $\left(n+1\right)!+2$].

Short answer

There exist $n$ consecutive composite integers.

Step-by-step solution

Step 1

DEFINITIONS

An integer$p$larger than$x$is called prime if the only positive factors of$p$are 1 and.

An integer that is not prime is called composite.

Step 2

Given: $n$ is a positive integer

To prove: There exist $n$ consecutive composite integers.

PROOF

The hint in the exercise prompt states that we should consider the $n$ consecutive integers starting with$\left(n+1\right)!+2$

Thus the $n$consecutive integers are then:

$$\begin{gathered} (n+1)!+2 \\ (n+1)!+3 \\ \cdots \\ (n+1)!+n \\ (n+1)!+(n+1) \end{gathered}$$

Since $(n+1)!=(n+1)\left(n\right)(n-1)\dots \left(2\right)\left(1\right)$,

$$\begin{gathered} 2\mid (n+1)! \\ 3\mid (n+1)! \\ \cdots \\ n\mid (n+1)! \\ (n+1)\mid (n+1)! \end{gathered}$$

An integer always divides itself.

$$\begin{gathered} 2\mid 2 \\ 3\mid 3 \\ \cdots \cdots \\ n\mid n \\ (n+1)\mid (n+1) \end{gathered}$$

Step 3

Finally, we know that if $k|a$and $k|b$, then $k|\left(a+b\right)$. Thus if $k|\left(n+1\right)!$and $k|k$, then $k|\left(n+1\right)!+k$.

$$\begin{gathered} 2\mid (n+1)!+2 \\ 3\mid (n+1)!+3 \\ \cdots \cdots \\ n\mid (n+1)!+n \\ (n+1)\mid (n+1)!+(n+1) \end{gathered}$$

We have then found a factor for each integer in the list of $n$ consecutive integers (while these factors are not 1 nor the integer itself), thus the $n$ consecutive integers are then all composite and thus we have found $n$consecutive composite integers.